Prove Angle APM = 1/2(Angle Q - Angle R)
Given : PM perpendicular to QR and AP is angle bisector of Angle QPR
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Step-by-step explanation:
Given: angle Q is greater than angle R and PA is the bisector of angle QPR and PM perpendicular QR
To Prove: angle APM = 1/2 (Q - R)
Proof: In triangle PQM
=> PMQ + MPQ + Q = 180 [Angle sum property of a triangle[
=> 90 + MPQ + Q = 180 (PMQ = 90)
=> Q = 90 - MPQ
=> 90 - MPQ = Q _____(1)
In triangle PMR
=> PMR + PRM + R = 180
=> 90 + MPR + R = 180
=> R = 90 - MPR
=> 90 - MPR = R _____(2)
Subtracting (1) and (2), we get
=> (90 - MPQ) - (90 - MPR) = Q - R
=> MPR - MPQ = Q - R
=> (RPA +APM) - (QPA - APM) = Q - R
=> QPA + APM -QPA + APM = Q - R [ As PA is bisector of QPR so, RPA=QPA]
=> 2 APM = Q - R
=> APM = 1/2 (Q - R)
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